Problem 20 - University Calculus: Early Transcendentals

Question

Find an antiderivative for each function. Do as many as you can mentally. Check your answers by differentiation. a. \(e^{-2 x}\) b. \(e^{4 x / 3}\) c. \(e^{-x / 5}\)

Step-by-Step Solution

Answer

Antiderivatives: a. \(-\frac{1}{2} e^{-2x} + C\), b. \(\frac{3}{4} e^{4x/3} + C\), c. \(-5 e^{-x/5} + C\)

Step 1: Understanding the Problem

We need to find the antiderivative of three exponential functions with different exponents. The key idea is to identify any changes of variables involved and apply the integration rules accordingly.

Step 2: Finding Antiderivative of Part a

The given function is \(e^{-2x}\). Recall the rule for antiderivatives: \[ \int e^{ax} \, dx = \frac{1}{a} e^{ax} + C \] where \(a\) is a constant.Here, \(a = -2\), so the antiderivative is: \[ \int e^{-2x} \, dx = -\frac{1}{2} e^{-2x} + C \]

Step 3: Check Part a by Differentiating

Differentiate \(-\frac{1}{2} e^{-2x} + C\) with respect to \(x\). Using the chain rule, we find:\[ \frac{d}{dx} \left(-\frac{1}{2} e^{-2x}\right) = -\frac{1}{2} \cdot (-2) e^{-2x} = e^{-2x} \]This matches the original function, confirming our antiderivative is correct.

Step 4: Finding Antiderivative of Part b

The given function is \(e^{4x/3}\). Using the antiderivative rule:\[ \int e^{ax} \, dx = \frac{1}{a} e^{ax} + C \]here \(a = \frac{4}{3}\). Thus, the antiderivative is:\[ \int e^{4x/3} \, dx = \frac{3}{4} e^{4x/3} + C \]

Step 5: Check Part b by Differentiating

Differentiate \(\frac{3}{4} e^{4x/3} + C\) with respect to \(x\). Applying the chain rule:\[ \frac{d}{dx} \left(\frac{3}{4} e^{4x/3}\right) = \frac{3}{4} \cdot \frac{4}{3} e^{4x/3} = e^{4x/3} \]This matches the original function, so the antiderivative is verified.

Step 6: Finding Antiderivative of Part c

The given function is \(e^{-x/5}\). Using the antiderivative rule:\[ \int e^{ax} \, dx = \frac{1}{a} e^{ax} + C \]here \(a = -\frac{1}{5}\). Thus, the antiderivative is:\[ \int e^{-x/5} \, dx = -5 e^{-x/5} + C \]

Step 7: Check Part c by Differentiating

Differentiate \(-5 e^{-x/5} + C\) with respect to \(x\). Using the chain rule:\[ \frac{d}{dx} (-5 e^{-x/5}) = -5 \cdot \left(-\frac{1}{5}\right) e^{-x/5} = e^{-x/5} \]This matches the original function, confirming our antiderivative is correct.

Key Concepts

Exponential Functions

Exponential functions are a key concept in calculus and mathematical analysis, often expressed in the form of \( e^{x} \). Here, "e" is the base of the natural logarithm, approximately equal to 2.718. These functions are powerful because they model growth and decay processes effectively.
When working with exponential functions in calculus, it's important to understand how changes in the exponent affect the function's behavior. In exercises involving antiderivatives of functions like \( e^{-2x} \), \( e^{4x/3} \), and \( e^{-x/5} \), understanding these variations is crucial. Generally, the exponential function \( e^{ax} \) will behave differently based on the constant \( a \).

• Positive \( a \): The function increases or grows.
• Negative \( a \): The function decreases or decays.

Recognizing these characteristics helps in determining the antiderivative, which is essentially the inverse operation to differentiation. This means that if you can find the rate of change of a function, you can also work backward to find its original form.

Integration Rules

Integration is the process of finding the antiderivative or integral of a function. It's a fundamental tool in calculus that encompasses several rules. The power of these rules lies in their ability to simplify complex expressions. Let's focus on the exponential function rules. The basic format is:
\[\int e^{ax} \, dx = \frac{1}{a} e^{ax} + C\]
This formula is derived from the property of exponential functions and the reverse of differentiation.

• \( C \) is the constant of integration, which is added to account for any constant that could have differentiated out to zero.
• \( a \) in the exponent significantly affects the integral and must be carefully managed, especially when \( a eq 1 \).

Understanding these integration rules allows you to solve problems involving exponential growth and decay efficiently.Even though the process might seem mechanical after practice, ensuring that you integrate correctly requires vigilance in applying the rules, especially under time pressure or during more complex exercises.

Chain Rule

The chain rule is a significant part of calculus used extensively in finding derivatives, particularly when dealing with composite functions. This rule is crucial for verifying antiderivatives as well.
The chain rule states that for a function \( f(g(x)) \), its derivative is:
\[(f(g(x)))' = f'(g(x)) \cdot g'(x)\]
This tells us to multiply the derivative of the outer function by the derivative of the inner function. When applied to exponential functions, like differentiating your antidifferentiated form, the chain rule checks if the operation returns to the original function.

• In part a, the function \( e^{-2x} \) undergoes differentiation in the solution to verify the correct antiderivative using the chain rule.
• This consistent checking via differentiation reassures that the found antiderivative is correctly determined.
• Similarly, verify your solutions by differentiating them to see if they return to their original state—this works as a built-in security check for mistakes.

Being comfortable with the chain rule ensures confidence in both taking derivatives and working back to find antiderivatives effectively.